# Page 149

**Location:** Chapter 14, bonus exercise 1, proposition (d)

## It is

(d) If \(Px(Pyz)\) sings on all days, so does \(P(Pxy)(Pyz)\).

## It should **disappear**

*[no sentence (d) to prove at all]*

## Short explanation

From the first three of Byrd's laws, proposition (d) **does not** follow.
(The rest of the exercise proceeds without problems, though.)

In the terms of Problem 1 of the Chapter, the particular arrangement

\[ x \notin S~~,~~~y \in S~~,~~~z \notin S \]acts as a counterexample, making it impossible to derive the fact that \(P(Pxy)(Pyz) \in S\) from \(Px(Pyz)\in S\) using the first three laws of Problem 1 alone.

In terms of the usual Aristotelian logic (i.e. under the isomorphism presented in "Discussion" on page 145), the following truth table makes the counterexample explicit:

\(x\) | \(y\) | \(z\) | \(y\to z\) | \(z\to y\) | \((x\to y)\to(y\to z)\) | \(x\to(y\to z)\) | \([x\to(y\to z)]\to[(x\to y)\to(y\to z)]\) |
---|---|---|---|---|---|---|---|

T | T | T | T | T | T | T | T |

T | T | F | F | T | F | F | T |

T | F | T | T | F | T | T | T |

T | F | F | T | F | T | T | T |

F | T | T | T | T | T | T | T |

F |
T |
F |
F |
T |
F |
T |
F |

F | F | T | T | T | T | T | T |

F | F | F | T | T | T | T | T |